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Rest energy question 
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#1
Aug1705, 02:43 AM

P: 740

Why is the rest energy mc^2 of an electron or positron 0.51 MeV?
Because if it's not moving, why is there "c"? 


#2
Aug1705, 10:17 AM

P: 576

E=mc^2. Plug in the electrons mass and divide my c^2. You get m=0.511 MeV/c^2.
c^2 is a constant of propotionality in this case and nothing more. 


#3
Aug1805, 02:24 AM

P: 76

its just the constant of the energy matter. no mroe it dont have to do with the speed.



#4
Aug1805, 03:26 AM

P: 740

Rest energy question
isn't c= the speed of light =3.0*10^8?



#5
Aug1805, 05:45 AM

P: 96




#6
Aug1805, 10:06 AM

P: 740

hmm... but i still think it's weird that you have to taked into account the speed of light a "rest" energy formula, because the electron's not moving...



#7
Aug1905, 08:54 AM

P: 96

You may find this link interesting:
http://www.pbs.org/wgbh/nova/einstein/ It explains stuff about the equation you were asking. ;) 


#8
Aug1905, 10:47 AM

P: 76

if u travel at 3*10^8m/s uŽll travel at ftl. the exact value of c is 299792458 m/s.



#9
Aug2005, 11:54 PM

P: 740

thanks! :)



#10
Aug2105, 02:14 PM

Emeritus
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P: 7,598

E^2  (pc)^2 = (mc^2)^2 This is a very valuable and useful relationship, but it requires that when p=0 that E is nonzero. We can substitute p=0 in the above equation and find E=mc^2 While we could write (E' + mc^2)  (pc)^2 = (mc^2)^2, where E' is the kinetic energy of the electron (excluding it's rest energy) it is much more convenient to make E include the rest energy of the electron. So that is what is done. The total energy of the electron (kinetic plus rest) also plays an important role in other aspects of physics, making the convention E=mc^2 when p=0 very useful, so useful that it is now a standard part of the definitions of physics. 


#11
Aug2105, 02:57 PM

P: 2,954

Pete 


#12
Aug2105, 05:14 PM

Emeritus
PF Gold
P: 8,147




#13
Aug2105, 11:00 PM

Emeritus
Sci Advisor
P: 7,598

I was thinking about my earlier response, and it''s fine as far as it goes but it's incomplete.
Probably the clearest physical experiment which demonstrates E=mc^2 is the creation of an electronpositron pair from gamma radiation, or the inverse process of the creation of gamma radiation from electronpositron annhiliaton. You find that the kinetic energy of the gamma rays generated (by annhilation) or needed (for pair creation) is E=mc^2 as a minimum. The minimum is achieved when the annhilating electronpositron pair (or the created pair) is at rest. It's also true as I remarked previously that E^2 = (mc^2)^2+(pc)^2 when E is defined as the total energy of the electron, i.e. E=(mc^2)+(kinetic energy), p is the momentum of the electron, and m is the rest mass (aka invariant mass) of the electron. 


#14
Aug2205, 02:49 AM

P: 2,954

"Why is the rest energy mc^2 of an electron or positron 0.51 MeV?" This question is easily rephrased by removing the clarification "mc^2" to give "Why is the rest energy an electron 0.51 MeV?" Classical relativity can't give that info. I don't see that the question was addressed at all here. The answer to his question about "c" is most easily given by providing a derivation and let them watch the "c" pop out. See http://www.geocities.com/physics_wor...ergy_equiv.htm But only asdf1 can tell us what is "plainly" the question. Pete 


#15
Aug2205, 04:13 AM

P: 740

@@a
i was wondering why rest energy has "c" in it's formula~ because "c" is usually referred to as the speed of light, so i thought that it was weird to have something that is moving included in a "rest" energy formula~ 


#16
Aug2205, 06:46 AM

P: 2,954

Pete 


#17
Aug2205, 08:09 AM

P: 1,017




#18
Aug2205, 10:18 AM

P: 740

thanks! :)
lol the username asdf1 rocks because it's easy to type... 


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